I got stuck working on a solved exercise about smooth maps. My opinion is that I lack some kind of basic geometry knowledge and I would really appreciate any suggestion in order to move on.
Let $M \subset \mathbb{R}^4$ be a 3-dimensional smooth submanifold of the Euclidean space ( it is obtained as a level set of a certain function, wich I won't mention). I am studying the smooth map $f:M \longrightarrow \mathbb{R}$ whose action on the domain is \begin{align*} M \ni x = (x^1, x^2,x^3,x^4) \longmapsto x^1-x^2. \end{align*} I have to find the points $p \in M$ for which the differential map of $f$ in $p$, \begin{align*} df_p:T_pM \longrightarrow T_{f(p)}\mathbb{R}, \end{align*} is not surjective.
As I was saying, this is a part of a solved exercise, but $i)$ I fear that my solution could contain flaws, $ii)$ I don't understand part of the "official" solution I have been given. Anyway, they are quite different.
$i)$ My solution was this one: I found that the matrix representation for $df_p$ is given by a constant matrix $(1,-1,0,0)$, which is independent of the point $p \in M$ and has always rank $1$. So there are no points $p$ for which $df_p$ is not surjective. Is this enough to conclude?
$ii)$ The provided solution is this one. Let $p=(x^1, x^2,x^3,x^4)$ be a point of $M$: then the space tangent to $M$ in $p$ is \begin{align*} T_pM = \{ a=(a^1,a^2,a^3,a^4) \in \mathbb{R}^4 | x^2a^1 + x^1a^2 + 2x^3a^3 + 2x^4a^4 = 0 \}, \end{align*} and the differential of $f$ in $p$ is \begin{align*} df_p: T_pM \ni a \longmapsto a^1-a^2 \in\mathbb{R}. \end{align*} Then $df_p$ is not surjective iff it is the constant zero map, iff $p$ is such that $a^1 - a^2 = 0$ for all $a \in T_pM$. Now the solutor says that, since the vectors $a \in T_pM$ must satisfy the equation $x^2a^1 + x^1a^2 + 2x^3a^3 + 2x^4a^4 = 0$, the points $p=(x^1, x^2,x^3,x^4)$ for which the differential is not surjective are those with $x^3=x^4=0$ and $x^1 + x^2 = 0$. I cannot understand why this is true.
You have $F: \mathbb{R}^4 \to \mathbb{R}, F(x^1,x^2,x^3,x^4) = x^1 - x^2,$ and $f = F \mid_M : M \to \mathbb{R}$. But $T_pM \ne T_p \mathbb{R}^4 = \mathbb{R}^4$. In fact, the equation $T_pM = \{ a=(a^1,a^2,a^3,a^4) \in \mathbb{R}^4 \mid x^2a^1 + x^1a^2 + 2x^3a^3 + 2x^4a^4 = 0 \}$ must come from a representation $M = g^{-1}(t)$ with a smooth map $g : \mathbb{R}^4 \to \mathbb{R}$ and a regular value $t$ of $g$. I also conclude that $p = 0$ is not contained in $M$, because otherwise $T_0 M = \mathbb{R}^4$ which is impossible.
This shows that your argument fails - you didn't take the correct tangent space.
Let $i : M \to \mathbb{R}^4$ denote the (smooth) inclusion map. $T_pi : T_pM \to T_p\mathbb{R}^4$ is simply the subspace inclusion. The chain rule gives $T_p f = T_p F \circ T_p i = T_p F \mid_{T_pM}$. This shows that $T_p f(a) = a^1 - a^2$ for $a \in T_p M$. I think it is then clear that $T_p f$ is not surjective iff $a^1 - a^2 = 0$ for all $a \in T_p M$. This is clearly satisfied if $x^3 = x^4 = 0$ and $x^1 + x^2 = 0$ (recall the constraint $x^2a^1 + x^1a^2 + 2x^3a^3 + 2x^4a^4 = 0$ and our above conclusion that $p \ne 0$ is not contained in $M$ which excludes $x^1 = x^2 = 0$ under the presence of $x^3 = x^4 = 0$). See Chris Custer's answer.
Conversely, assume $a^1 - a^2 = 0$ for all $a \in T_p M$. That is, $T_pM = \{ a=(a^1,a^2,a^3,a^4) \in \mathbb{R}^4 \mid a^1 = a^2, (x^1 + x^2)a^2 + 2x^3a^3 + 2x^4a^4 = 0 \}$. Define $r : T_pM \to V = \{ (a^2,a^3,a^4) \in \mathbb{R}^3 \mid (x^1 + x^2)a^2 + 2x^3a^3 + 2x^4a^4 = 0 \}$ by $r(a^1,a^2,a^3,a^4) =(a^2,a^3,a^4)$. It is readily seen that this is an isomorphism of vector spaces. Since $T_p M$ has dimension $3$, also $V$ must have dimension $3$ and we conclude $V = \mathbb{R}^3$. This is only possible when $x^1 + x^2 = x^3 = x^4 = 0$.
The set of these $p = (x^1,x^2,x^3,x^4)$ is $3$-dimensional subspace $H$ of $\mathbb{R}^4$ and $H \cap M$ is the set on which $T_pf$ is not surjective.